Solution-Manual-for-Physics-2nd-Edition-by-Giambattista

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Chapter 1
INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many study physics for the sense of fulfillment that comes with learning about
the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. (a) 3
(b) 9
5. Scientific notation eliminates the need to write many zeros in very large or small numbers. Also, the appropriate
number of significant digits is unambiguous when written this way.
6. In scientific notation the decimal point is placed after the first (leftmost) numeral. The number of digits written
equals the number of significant figures.
7. Not all of the significant digits are precisely known. The least significant digit (rightmost) is an estimate and is
less precisely known than the others.
8. It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is
known and not mislead the reader by writing digits that are not at all known to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI system.
10. The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in
terms of these units and their powers of ten. The U.S. Customary system contains units that are primarily of
historical origin and are not based upon powers of ten. As a result of this international acceptance and the ease of
manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system.
11. Fathoms, kilometers, miles, and inches are units with dimensions of length. Grams and kilograms are units with
dimensions of mass. Years, months, and seconds are units with dimensions of time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
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15. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. It may
also be useful to explore other possible methods of solution as a check on the validity of the first.
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence.
1.37 &times; 1.8 m = 2.5 m
2. Strategy There are
60 s 60 min 24 h
&times;
&times;
= 86, 400 seconds in one day and 24 hours in one day.
1 min
1h
1d
Solution Find the ratio of the number of seconds in a day to the number of hours in a day.
86, 400 24 &times; 3600
=
= 3600 1
24
24
3. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 1.160S1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 1.160
⎝ r1 ⎠
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
4. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 2.0 S1 = 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 2.0
⎝ r1 ⎠
r2
= 2.0 = 1.4
r1
The radius of the balloon increases by a factor of 1.4.
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5. Strategy The surface area S and the volume V are given by S = 6s 2 and V = s3 , respectively.
Solution Find the ratio of the surface area to the volume.
S 6s 2
6
=
=
3
V
s
s
6. Strategy To find the factor Samantha’s height increased, divide her new height by her old height. Subtract 1 from
this value and multiply by 100 to find the percent increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length
1
map area
=
= 10−4 , so
= (10−4 ) 2 = 10−8 .
actual length 10, 000
actual area
8. Strategy Let X be the original value of the index.
Solution Find the net percentage change in the index for the two days.
(first day change) &times; (second day change) = [ X &times; (1 + 0.0500)] &times; (1 − 0.0500) = 0.9975 X
The net percentage change is (0.9975 − 1) &times; 100% = −0.25%, or down 0.25% .
9. Strategy Use a proportion.
Solution Find Jupiter’s orbital period.
T 2 R3
T 2 ∝ R3 , so J = J = 5.193. Thus, TJ = 5.193/2 TE = 11.8 yr .
TE2 RE3
10. Strategy The area of the circular garden is given by A = π r 2 . Let the original and final areas be A1 = π r12 and
A2 = π r22 , respectively.
Solution Calculate the percentage increase of the area of the garden plot.
π r 2 − π r12
r2 − r2
1.252 r12 − r12
∆A
1.252 − 1
&times; 100% = 2
&times; 100% = 2 1 &times; 100% =
&times; 100% =
&times; 100% = 56%
A
1
π r12
r12
r12
11. Strategy The area of the poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 = 2 w2 = (0.800 1 )(0.800w1 ) = 0.640 1w1 = 0.640 A1
A1 − A2
A − 0.640 A1
&times; 100% = 1
&times; 100% = 36.0%
A1
A1
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12. Strategy The volume of the rectangular room is given by V = wh. Let the original and final volumes be
V1 = 1w1h1 and V2 = 2 w2 h2 , respectively.
Solution Find the factor by which the volume of the room increased.
V2
w h
(1.50 1 )(2.00 w1 )(1.20h1 )
= 2 2 2 =
= 3.60
V1
1w1h1
1w1h1
13. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer
with the number of significant figures determined by the less precise of the two numbers.
Solution Perform the operation with the appropriate number of significant figures.
3.783 &times; 106 kg + 1.25 &times; 108 kg = 0.03783 &times; 108 kg + 1.25 &times; 108 kg = 1.29 &times; 108 kg
(b) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Perform the operation with the appropriate number of significant figures.
(3.783 &times; 106 m) &divide; (3.0 &times; 10−2 s) = 1.3 &times; 108 m s
14. (a) Strategy Move the decimal point eight places to the left and multiply by 108.
Solution Write the number in scientific notation.
290,000,000 people = 2.9 &times; 108 people
(b) Strategy Move the decimal point 15 places to the right and multiply by 10−15.
Solution Write the number in scientific notation.
0.000 000 000 000 003 8 m = 3.8 &times; 10−15 m
15. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then subtract and give the
answer with the number of significant figures determined by the less precise of the two numbers.
Solution Perform the calculation using an appropriate number of significant figures.
3.68 &times; 107 g − 4.759 &times; 105 g = 3.68 &times; 107 g − 0.04759 &times; 107 g = 3.63 &times; 107 g
(b) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Perform the calculation using an appropriate number of significant figures.
6.497 &times; 104 m 2
= 1.273 &times; 102 m
5.1037 &times; 102 m
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16. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer
with the number of significant figures determined by the less precise of the two numbers.
Solution Write your answer using the appropriate number of significant figures.
6.85 &times; 10−5 m + 2.7 &times; 10−7 m = 6.85 &times; 10−5 m + 0.027 &times; 10−5 m = 6.88 &times; 10−5 m
(b) Strategy Add and give the answer with the number of significant figures determined by the less precise of
the two numbers.
Solution Write your answer using the appropriate number of significant figures.
702.35 km + 1897.648 km = 2600.00 km
(c) Strategy Multiply and give the answer with the number of significant figures determined by the number with
the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
5.0 m &times; 4.3 m = 22 m 2
(d) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
( 0.04 π ) cm =
0.01 cm
(e) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
( 0.040 π ) m =
0.013 m
17. Strategy Multiply and give the answer in scientific notation with the number of significant figures determined by
the number with the fewest significant figures.
Solution Solve the problem.
(3.2 m) &times; (4.0 &times; 10−3 m) &times; (1.3 &times; 10−8 m) = 1.7 &times; 10−10 m3
18. Strategy Follow the rules for identifying significant figures.
Solution
(a) All three digits are significant, so 7.68 g has 3 significant figures.
(b) The first zero is not significant, since it is used only to place the decimal point. The digits 4 and 2 are
significant, as is the final zero, so 0.420 kg has 3 significant figures.
(c) The first two zeros are not significant, since they are used only to place the decimal point. The digits 7 and 3
are significant, so 0.073 m has 2 significant figures.
(d) All three digits are significant, so 7.68 &times; 105 g has 3 significant figures.
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(e) The zero is significant, since it comes after the decimal point. The digits 4 and 2 are significant as well, so
4.20 &times; 103 kg has 3 significant figures.
(f) Both 7 and 3 are significant, so 7.3 &times; 10−2 m has 2 significant figures.
(g) Both 2 and 3 are significant. The two zeros are significant as well, since they come after the decimal point, so
2.300 &times; 104 s has 4 significant figures.
19. Strategy Divide and give the answer with the number of significant figures determined by the number with the
fewest significant figures.
Solution Solve the problem.
3.21 m
3.21 m
=
= 459 m s
7.00 ms 7.00 &times; 10−3 s
20. Strategy Convert each length to meters. Then, rewrite the numbers so that the power of 10 is the same for each.
Finally, add and give the answer with the number of significant figures determined by the less precise of the two
numbers.
Solution Solve the problem.
3.08 &times; 10−1 km + 2.00 &times; 103 cm = 3.08 &times; 102 m + 2.00 &times; 101 m = 3.08 &times; 102 m + 0.200 &times; 102 m = 3.28 &times; 102 m
21. Strategy There are approximately 39.37 inches per meter.
Solution Find the thickness of the cell membrane in inches.
7.0 &times; 10−9 m &times; 39.37 inches m = 2.8 &times; 10−7 inches
22. (a) Strategy There are approximately 3.785 liters per gallon and 128 ounces per gallon.
Solution Find the number of fluid ounces in the bottle.
128 fl oz
1 gal
1L
&times;
&times; 355 mL &times;
= 12.0 fluid ounces
1 gal
3.785 L
103 mL
(b) Strategy From part (a), we have 355 mL = 12.0 fluid ounces.
Solution Find the number of milliliters in the drink.
355 mL
16.0 fl oz &times;
= 473 mL
12.0 fl oz
23. Strategy There are approximately 3.281 feet per meter.
Solution Convert to meters and identify the order of magnitude.
(a) 1595.5 ft &times;
(b) 6016 ft &times;
1m
= 4.863 &times; 102 m ; the order of magnitude is 102 .
3.281 ft
1m
= 1.834 &times; 103 m ; the order of magnitude is 103 .
3.281 ft
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24. Strategy There are 3600 seconds in one hour and 1000 m in one kilometer.
Solution Convert 1.00 kilometers per hour to meters per second.
1.00 km
1h
1000 m
&times;
&times;
= 0.278 m s
1h
3600 s 1 km
25. (a) Strategy There are 60 seconds in one minute, 5280 feet in one mile, and 3.28 feet in one meter.
Solution Express 0.32 miles per minute in meters per second.
0.32 mi 1 min 5280 ft
1m
&times;
&times;
&times;
= 8.6 m s
1 min
60 s
1 mi
3.28 ft
(b) Strategy There are 60 minutes in one hour.
Solution Express 0.32 miles per minute in miles per hour.
0.32 mi 60 min
&times;
= 19 mi h
1 min
1h
26. Strategy There are 0.6214 miles in 1 kilometer.
Solution Find the length of the marathon race in miles.
0.6214 mi
= 26.22 mi
42.195 km &times;
1 km
27. Strategy Calculate the change in the exchange rate and divide it by the original price to find the drop.
Solution Find the actual drop in the value of the dollar over the first year.
1.27 − 1.45 −0.18
=
= −0.12
1.45
1.45
The actual drop is 0.12 or 12% .
28. Strategy There are 1000 watts in one kilowatt and 100 centimeters in one meter.
Solution Convert 1.4 kW m 2 to W cm 2 .
2
1.4 kW 1000 W ⎛ 1 m ⎞
2
&times;
&times;⎜
⎟ = 0.14 W cm
1 kW ⎝ 100 cm ⎠
1 m2
29. Strategy There are 1000 grams in one kilogram and 100 centimeters in one meter.
Solution Find the density of mercury in units of g cm3 .
3
1.36 &times; 104 kg 1000 g ⎛ 1 m ⎞
3
&times;
&times;⎜
⎟ = 13.6 g cm
1 kg ⎝ 100 cm ⎠
1 m3
30. Strategy The distance traveled d is equal to the rate of travel r times the time of travel t. There are 1000
milliseconds in one second.
Solution Find the distance the molecule would move.
459 m
1s
d = rt =
&times; 7.00 ms &times;
= 3.21 m
1s
1000 ms
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31. Strategy There are 1000 meters in a kilometer and 1,000,000 millimeters in a kilometer.
Solution Find the product and express the answer in km3 with the appropriate number of significant figures.
1 km
1 km
(3.2 km) &times; (4.0 m) &times; (13 &times; 10−3 mm) &times;
&times;
= 1.7 &times; 10−10 km3
1000 m 1,000,000 mm
32. (a) Strategy There are 12 inches in one foot and 2.54 centimeters in one inch.
Solution Find the number of square centimeters in one square foot.
2
2
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞
2
1 ft 2 &times; ⎜
⎟ &times;⎜
⎟ = 929 cm
1
ft
1
in
⎝
⎠ ⎝
⎠
(b) Strategy There are 100 centimeters in one meter.
Solution Find the number of square centimeters in one square meter.
2
⎛ 100 cm ⎞
4
2
1 m2 &times; ⎜
⎟ = 1&times; 10 cm
⎝ 1m ⎠
(c) Strategy Divide one square meter by one square foot. Estimate the quotient.
Solution Find the approximate number of square feet in one square meter.
1 m 2 10, 000 cm 2
=
≈ 11
1 ft 2
929 cm 2
33. (a) Strategy There are 12 inches in one foot, 2.54 centimeters in one inch, and 60 seconds in one minute.
Solution Express the snail’s speed in feet per second.
5.0 cm 1min
1 in
1 ft
&times;
&times;
&times;
= 2.7 &times; 10−3 ft s
1 min 60 s 2.54 cm 12 in
(b) Strategy There are 5280 feet in one mile, 12 inches in one foot, 2.54 centimeters in one inch, and 60 minutes
in one hour.
Solution Express the snail’s speed in miles per hour.
5.0 cm 60 min
1 in
1 ft
1 mi
&times;
&times;
&times;
&times;
= 1.9 &times; 10−3 mi h
1 min
1h
2.54 cm 12 in 5280 ft
34. Strategy A micrometer is 10−6 m and a millimeter is 10−3 m; therefore, a micrometer is 10−6 10−3 = 10−3 mm.
Solution Find the area in square millimeters.
2
⎛ 10−3 mm ⎞
150 &micro;m 2 &times; ⎜
= 1.5 &times; 10−4 mm 2
⎜ 1 &micro;m ⎟⎟
⎝
⎠
35. Strategy Replace each quantity in U = mgh with its SI base units.
Solution Find the combination of SI base units that are equivalent to joules.
U = mgh ⇒ J = kg &times; m s 2 &times; m = kg ⋅ m 2 ⋅ s −2
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36. (a) Strategy Replace each quantity in ma and kx with its dimensions.
Solution Show that the dimensions of ma and kx are equivalent.
[L]
[M]
[L]
ma has dimensions [M] &times;
&times; [L] = [M] &times;
.
and kx has dimensions
2
2
[T]
[T]
[T]2
Since [M][L][T]−2 = [M][L][T]−2 , the dimensions are equivalent.
(b) Strategy Use the results of part (a).
Solution Since F = ma and F = − kx, the dimensions of the force unit are [M][L][T]−2 .
37. Strategy Replace each quantity in T 2 = 4π 2 r 3 (GM ) with its dimensions.
Solution Show that the equation is dimensionally correct.
4π 2 r 3
[L]3
[L]3 [M][T]2
has dimensions
=
&times;
= [T]2 .
T 2 has dimensions [T]2 and
3
3
[L]
[M]
GM
[L]
&times; [M]
[M][T]2
Since [T]2 = [T]2 , the equation is dimensionally correct.
38. Strategy Determine the SI unit of momentum using a process of elimination.
Solution Find the SI unit of momentum.
p2
kg ⋅ m 2
kg 2 ⋅ m 2
. Since the SI unit for m is kg, the SI unit for p 2 is
. Taking the square
has units of
K=
2
2m
s
s2
root, we find that the SI unit for momentum is kg ⋅ m ⋅ s −1 .
39. (a) Strategy Replace each quantity (except for V) in FB = ρ gV with its dimensions.
Solution Find the dimensions of V.
F
[MLT −2 ]
V = B has dimensions
= [L3 ] .
−3
−2
ρg
[ML ] &times; [LT ]
(b) Strategy and Solution Since velocity has dimensions [LT −1 ] and volume has dimensions [L3 ], the correct
interpretation of V is that is represents volume .
40. Strategy Replace v, r, ω , and m with their dimensions. Then use dimensional analysis to determine how v
depends upon some or all of the other quantities.
[L]
1
, [L],
, and [M], respectively. No combination of r, ω , and m
[T]
[T]
1 [L]
=
and there is no dimensionless
gives dimensions without [M], so v does not depend upon m. Since [L] &times;
[T] [T]
Solution v, r , ω , and m have dimensions
constant involved in the relation, v is equal to the product of ω and r , or v = ω r .
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41. Strategy Approximate the distance from your eyes to a book held at your normal reading distance.
Solution The normal reading distance is about 30-40 cm, so the approximate distance from your eyes to a book
you are reading is 30-40 cm.
42. Strategy Estimate the length, width, and height of your textbook. Then use V = wh to estimate its volume.
Solution Find the approximate volume of your physics textbook in cm3 .
The length, width, and height of your physics textbook are approximately 30 cm, 20 cm, and 4.0 cm, respectively.
V = wh = (30 cm)(20 cm)(4.0 cm) = 2400 cm3
43. (a) Strategy and Solution The mass of the lower leg is about 5 kg and that of the upper leg is about 7 kg, so an
order of magnitude estimate of the mass of a person’s leg is 10 kg.
(b) Strategy and Solution The length of a full size school bus is greater than 1 m and less than 100 m, so an
order of magnitude estimate of the length of a full size school bus is 10 m.
44. Strategy and Solution A normal heart rate is about 70 beats per minute and a person lives for about 70 years, so
70 beats
70 y
5.26 &times; 105 min
&times;
&times;
= 2.6 &times; 109 times per lifetime, or about 3 &times; 109 .
the heart beats about
1 min
lifetime
1y
45. Strategy (Answers will vary.) In this case, we use San Francisco, CA for the city. The population of San
Francisco is approximately 750,000. Assume that there is one automobile for every two residents of San
Francisco, that an average automobile needs three repairs or services per year, and that the average shop can
service 10 automobiles per day.
Solution Estimate the number of automobile repair shops in San Francisco.
3 repairs 1 y
0.01 repairs
.
&times;
≈
If an automobile needs three repairs or services per year, then it needs
auto ⋅ y 365 d
auto ⋅ d
1 auto
&times; 750, 000 residents ≈ 4 &times; 105 autos.
2 residents
If a shop requires one day to service 10 autos, then the number of shops-days per repair is
1d
0.1 shop ⋅ d
=
1 shop &times;
.
10 repairs
repair
If there is one auto for every two residents, then there are
0.01 repairs 0.1 shop ⋅ d
&times;
= 400 shops .
auto ⋅ d
repair
Checking the phone directory, we find that there are approximately 463 automobile repair and service shops in
400 − 463
&times; 100% = −16% . The estimate was 16% too low, but in the ball
San Francisco. The estimate is off by
400
park!
The estimated number of auto shops is 4 &times; 105 autos &times;
46. Strategy Estimate the appropriate orders of magnitude.
Solution Find the order of magnitude of the number of seconds in one year.
seconds/minute ~ 102
minutes/hour ~ 102
hours/day ~ 101
days/year ~ 102
102 ⋅102 ⋅101 ⋅102 = 107 s
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Chapter 1: Introduction
47. Strategy One story is about 3 m high.
Solution Find the order of magnitude of the height in meters of a 40-story building.
(3 m)(40) ~ 100 m
48. Strategy To determine if c and A0 are correct, graph A versus B3 .
49. Strategy The plot of temperature versus
elapsed time is shown. Use the graph to
answer the questions.
A on the vertical axis and B3 on the horizontal axis .
103.00
Temperature (&deg;F)
Solution To graph A versus B3 , graph
102.00
101.00
100.00
10 A.M.
11 A.M. 12 P.M.
Time
1 P.M.
Solution
(a) By inspection of the graph, it appears that the temperature at noon was 101.8&deg;F.
(b) Estimate the slope of the line.
102.6&deg;F − 100.0&deg;F
2.6&deg;F
=
= 0.9 &deg;F h
m=
1:00 P.M. − 10:00 A.M.
3h
(c) In twelve hours, the temperature would, according to the trend, be approximately
T = (0.9 &deg;F h)(12 h) + 102.5&deg;F = 113&deg;F.
The patient would be dead before the temperature reached this level. So, the answer is no.
50. Strategy Use the slope-intercept form, y = mx + b.
Solution Since x is on the vertical axis, it corresponds to y. Since t 4 is on the horizontal axis, it corresponds to x
(in y = mx + b) . So, the equation for x as a function of t is x = (25 m s 4 )t 4 + 3 m .
51. Strategy Use the two temperatures and their corresponding times to find the rate of temperature change with
respect to time (the slope of the graph of temperature vs. time). Then, write the linear equation for the temperature
with respect to time and find the temperature at 3:35 P.M.
Solution Find the rate of temperature change.
∆T 101.0&deg;F − 97.0&deg;F
=
= 1.0 &deg;F h
m=
∆t
4.0 h
Use the slope-intercept form of a graph of temperature vs. time to find the temperature at 3:35 P.M.
T = mt + T0 = (1.0 &deg;F h)(3.5 h) + 101.0&deg;F = 104.5&deg;F
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Strategy Plot the weights and ages on a
weight versus age graph.
Solution See the graph.
20.0
Weight (lb)
52. (a)
Physics
15.0
10.0
5.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0
Age in months
(b) Strategy Find the slope of the best-fit line between age 0.0 and age 5.0 months.
Solution Find the slope.
13.6 lb − 6.6 lb
7.0 lb
=
= 1.4 lb mo
m=
5.0 mo − 0.0 mo 5.0 mo
(c) Strategy Find the slope of the best-fit line between age 5.0 and age 10.0 months.
Solution Find the slope.
17.5 lb − 13.6 lb
3.9 lb
=
= 0.78 lb mo
m=
10.0 mo − 5.0 mo 5.0 mo
(d) Strategy Write a linear equation for the weight of the baby as a function of time. The slope is that found in
part (b), 1.4 lb mo. The intercept is the weight of the baby at five months of age.
Solution Find the projected weight of the child at age 12.
W = (1.4 lb mo)(144 mo − 5 mo) + 13.6 lb = 210 lb
53. Strategy Put the equation that describes the line in slope-intercept form, y = mx + b.
at = v − v0
v = at + v0
Solution
(a) v is the dependent variable and t is the independent variable, so a is the slope of the line.
(b) The slope-intercept form is y = mx + b. Find the vertical-axis intercept.
v ↔ y, t ↔ x, a ↔ m, so v0 ↔ b.
Thus, +v0
is the vertical-axis intercept of the line.
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54. (a) Strategy The equation of the speed versus time is given by v = at + v0 , where a = 6.0 m s 2 and
v0 = 3.0 m s.
Solution Find the change in speed.
v2 = at2 + v0
− (v1 = at1 + v0 )
v2 − v1 = a (t2 − t1 )
v2 − v1 = (6.0 m s 2 )(6.0 s − 4.0 s) = 12 m s
(b) Strategy Use the equation found in part (a).
Solution Find the speed when the elapsed time is equal to 5.0 seconds.
v = (6.0 m s 2 )(5.0 s) + 3.0 m s = 33 m s
55. (a) Strategy Plot the decay rate on the vertical axis and the time on the horizontal axis.
Solution The plot is shown.
Decay Rate (decays/s)
450
300
150
0
0 10 20 30 40 50 60 70 80 90 100
Time (min)
(b) Strategy Plot the natural logarithm of the decay rate on the vertical axis and the time on the horizontal axis.
Natural Logarithm of the
Decay Rate
Solution The plot is shown.
Presentation of the data in this form—as the
natural logarithm of the decay rate—might be
useful because the graph is linear.
6.0
4.0
2.0
0
0 10 20 30 40 50 60 70 80 90 100
Time (min)
56. (a) Strategy Refer to the figure. Use the definition of the slope of a line and the fact that the vertical axis
intercept is the x-value corresponding to t = 0.
Solution Compute the slope.
∆x 17.0 km − 3.0 km
=
= 1.6 km h .
∆t
9.0 h − 0.0 h
When t = 0, x = 3.0 km; therefore, the vertical axis intercept is 3.0 km.
(b) Strategy and Solution The physical significance of the slope of the graph is that it represents the speed of
the object. The physical significance of the vertical axis intercept is that it represents the starting position of
the object (position at time zero).
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57. Strategy For parts (a) through (d), perform the calculations.
Solution
(a) 186.300 + 0.0030 = 186.303
(b) 186.300 − 0.0030 = 186.297
(c) 186.300 &times; 0.0030 = 0.56
(d) 186.300 0.0030 = 62, 000
(e) Strategy For cases (a) and (b), the percent error is given by
0.0030
&times; 100%.
Actual Value
Solution Find the percent error.
0.0030
&times; 100% = 0.0016%
Case (a):
186.303
0.0030
&times; 100% = 0.0016%
Case (b):
186.297
For case (c), ignoring 0.0030 causes you to multiply by zero and get a zero result. For case (d), ignoring
0.0030 causes you to divide by zero.
(f) Strategy Make a rule about neglecting small values using the results obtained above.
Solution
You can neglect small values when they are added to or subtracted from sufficiently large values.
The term “sufficiently large” is determined by the number of significant figures required.
58. Strategy The weight is proportional to the mass and inversely proportional to the square of the radius, so
W ∝ m r 2 . Thus, for Earth and Jupiter, we have WE ∝ mE rE2 and WJ ∝ mJ rJ2 .
Solution Form a proportion.
WJ
WE
=
mJ rJ2
mE rE2
=
mJ ⎛ rE
⎜
mE ⎜⎝ rJ
2
⎞
320mE
⎟⎟ =
mE
⎠
On Jupiter, the apple would weigh
⎛ rE
⎜⎜
⎝ 11rE
320 (1.0
121
2
⎞
320
⎟⎟ =
121
⎠
N) = 2.6 N .
59. Strategy Assuming that the cross section of the artery is a circle, we use the area of a circle, A = π r 2.
Solution
A1 = π r12 and A2 = π r22 = π (2.0r1 )2 = 4.0π r12 .
Form a proportion.
A2 4.0π r12
=
= 4.0
A1
π r12
The cross-sectional area of the artery increases by a factor of 4.0.
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Chapter 1: Introduction
60. (a) Strategy The diameter of the xylem vessel is one six-hundredth of the magnified image.
Solution Find the diameter of the vessel.
d magnified 3.0 cm
d actual =
=
= 5.0 &times; 10−3 cm
600
600
(b) Strategy The area of the cross section is given by A = π r 2 = π (d 2)2 = (1 4)π d 2 .
Solution Find by what factor the cross-sectional area has been increased in the micrograph.
Amagnified 14 π d magnified 2 ⎛ 3.0 cm ⎞2
=
=⎜
= 360,000 .
2
−3 cm ⎟
1πd
Aactual
&times;
5.0
10
⎝
⎠
actual
4
61. Strategy If s is the speed of the molecule, then s ∝ T where T is the temperature.
Solution Form a proportion.
Tcold
scold
=
swarm
Twarm
Find scold .
scold = swarm
Tcold
250.0 K
= (475 m s)
= 434 m s
Twarm
300.0 K
62. Strategy Use dimensional analysis to convert from furlongs per fortnight to the required units.
Solution
(a) Convert to &micro;m s.
1 furlong
220 yd 1 fortnight
1 day
3 ft
1 m 1, 000, 000 &micro;m
&times;
&times;
&times;
&times;
&times;
&times;
= 166 &micro;m s
1 fortnight 1 furlong 14 days
86,400 s 1 yd 3.28 ft
1m
(b) Convert to km day.
1 furlong
220 yd 1 fortnight 3 ft
1m
1 km
&times;
&times;
&times;
&times;
&times;
= 0.0144 km day
1 fortnight 1 furlong 14 days 1 yd 3.28 ft 1000 m
63. Strategy Use the rules for determining significant figures and for writing numbers in scientific notation.
Solution
(a) 0.00574 kg has three significant figures, 5, 7, and 4. The zeros are not significant, since they are used only to
place the decimal point. To write this measurement in scientific notation, we move the decimal point three
places to the right and multiply by 10−3.
(b) 2 m has one significant figure, 2. This measurement is already written in scientific notation
(c) 0.450 &times; 10−2 m has three significant figures, 4, 5, and the 0 to the right of 5. The zero is significant, since it
comes after the decimal point and is not used to place the decimal point. To write this measurement in
scientific notation, we move the decimal point one place to the right and multiply by 10−1.
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(d) 45.0 kg has three significant figures, 4, 5, and 0. The zero is significant, since it comes after the decimal point
and is not used to place the decimal point. To write this measurement in scientific notation, we move the
decimal point one place to the left and multiply by 101.
(e) 10.09 &times; 104 s has four significant figures, 1, 9, and the two zeros. The zeros are significant, since they are
between two significant figures. To write this measurement in scientific notation, we move the decimal point
one place to the left and multiply by 101.
(f) 0.09500 &times; 105 mL has four significant figures, 9, 5, and the two zeros to the right of 5. The zeros are
significant, since they come after the decimal point and are not used to place the decimal point. To write this
measurement in scientific notation, we move the decimal point two places to the right and multiply by 10−2.
The results of parts (a) through (f) are shown in the table below.
Measurement
Significant Figures
Scientific Notation
(a)
0.00574 kg
3
5.74 &times; 10−3 kg
(b)
2m
1
2m
(c)
0.450 &times; 10−2 m
3
4.50 &times; 10−3 m
(d)
45.0 kg
3
4.50 &times; 101 kg
(e)
10.09 &times; 104 s
4
1.009 &times; 105 s
(f)
0.09500 &times; 105 mL
4
9.500 &times; 103 mL
64. Strategy Use the conversion factors from the inside cover of the book.
Solution
(a)
12.5 US gal 3.785 L 103 mL 0.06102 in 3
&times;
&times;
&times;
= 2890 in 3
1
US gal
L
mL
(b)
2887 in 3 ⎛ 1 cubit ⎞
&times;⎜
⎟ = 0.495 cubic cubits
1
⎝ 18 in ⎠
3
65. Strategy Use the metric prefixes n (10−9 ), &micro; (10−6 ), m (10−3 ), or M (106 ).
Solution
(a) M (or mega) is equal to 106 , so 6 &times; 106 m = 6 Mm .
(b) There are approximately 3.28 feet in one meter, so 6 ft &times;
1m
= 2m .
3.28 ft
(c) &micro; (or micro) is equal to 10−6 , so 10−6 m = 1 &micro;m .
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(d) n (or nano) is equal to 10−9 , so 3 &times; 10−9 m = 3 nm .
(e) n (or nano) is equal to 10−9 , so 3 &times; 10−10 m = 0.3 nm .
66. Strategy The volume of the spherical virus is given by Vvirus = (4 3)π rvirus3 . The volume of viral particles is one
billionth the volume of the saliva.
Solution Calculate the number of viruses that have landed on you.
10−9 Vsaliva
0.010 cm3
number of viral particles =
=
= 104 viruses
3 10−7 cm 3
Vvirus
9
85
nm
10 43 π
2
1 nm
( )(
)
(
)
67. Strategy The circumference of a viroid is approximately 300 times 0.35 nm. The diameter is given by C = π d ,
or d = C π .
Solution Find the diameter of the viroid in the required units.
(a) d =
300(0.35 nm) 10−9 m
&times;
= 3.3 &times; 10−8 m
1 nm
π
(b) d =
300(0.35 nm) 10−3 &micro;m
&times;
= 3.3 &times; 10−2 &micro;m
1 nm
π
(c) d =
300(0.35 nm) 10−7 cm
1 in
&times;
&times;
= 1.3 &times; 10−6 in
1 nm
2.54 cm
π
68. (a) Strategy There are 3.28 feet in one meter.
Solution Find the length in meters of the largest recorded blue whale.
1m
1.10 &times; 102 ft &times;
= 33.5 m
3.28 ft
(b) Strategy Divide the length of the largest recorded blue whale by the length of a double-decker London bus.
Solution Find the length of the blue whale in double-decker-bus lengths.
1.10 &times; 102 ft
1m
&times;
= 4.2 bus lengths
m
3.28 ft
8.0
bus length
69. Strategy The volume of the blue whale can be found by dividing the mass of the whale by its average density.
Solution Find the volume of the blue whale in cubic meters.
V=
m
ρ
=
1.9 &times; 105 kg
3
1000 g ⎛ 1 m ⎞
2 3
&times;⎜
⎟ = 2.2 &times; 10 m
3
1
kg
100
cm
⎝
⎠
0.85 g cm
&times;
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70. Strategy The shape of a sheet of paper (when not deformed) is a rectangular prism. The volume of a rectangular
prism is equal to the product of its length, width, and height (or thickness).
Solution Find the volume of a sheet of paper in cubic meters.
1m
0.0254 m
1m
&times;
&times;
= 6.0 &times; 10−6 m3
27.95 cm &times; 8.5 in &times; 0.10 mm &times;
100 cm
1 in
1000 mm
71. (a) Strategy a has dimensions
[L]
[T]2
[L]
; v has dimensions [T] ; r has dimension [L].
2
[L]2 1
[L]
Solution If we square v and divide by r, we have vr , which implies that
⋅ =
, which are the
[T]2 [L] [T]2
dimensions for a. Therefore, we can write a = K
v2
r
, where K is a dimensionless constant.
(b) Strategy Divide the new acceleration by the old, and use the fact that the new speed is 1.100 times the old.
Solution Find the percent increase in the radial acceleration.
v2
2
2
2
⎛ 1.100v1 ⎞
a2 K r ⎛ v2 ⎞
=
=
=
= 1.1002 = 1.210
⎜
⎟
⎜
⎟
a1 K v12 ⎝ v1 ⎠
⎝ v1 ⎠
r
1.210 − 1 = 0.210, so the radial acceleration increases by 21.0%.
72. Strategy Replace each quantity in v = K λ p g q by its units. Then, use the relationships between p and q to
determine their values.
Solution Find the values of p and q.
m
mq m p + q
.
= mp ⋅
=
In units,
s
s 2q
s2q
So, we have the following restrictions on p and q: p + q = 1 and 2q = 1.
Solve for q and p.
p + q =1
2q = 1
1
p + =1
1
q=
2
2
1
p=
2
Thus, v = K λ1 2 g1 2 = K λ g .
73. Strategy There are 2.54 cm in one inch and 3600 seconds in one hour.
Solution Find the conversion factor for changing meters per second to miles per hour.
1 m 100 cm
1 in
1 ft
1 mi
3600 s
&times;
&times;
&times;
&times;
&times;
= 2.24 mi h = 1 m s
1s
1m
2.54 cm 12 in 5280 ft
1h
So, for a quick, approximate conversion, multiply by 2.
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Chapter 1: Introduction
74. Strategy The order of magnitude of the volume of water required to fill a bathtub is 101 ft 3 . The order of
magnitude of the number of cups in a cubic foot is 102.
Solution Find the order of magnitude of the number of cups of water required to fill a bathtub.
101 ft 3 &times; 102 cups ft 3 = 103 cups
75. Strategy Since there are hundreds of millions of people in the U.S., a reasonable order-of-magnitude estimate of
the number of automobiles is 108. There are about 365 days per year; that is, about 102. A reasonable estimate of
the gallons used per day per person is greater than one, but less than one hundred; that is, 101.
Solution Calculate the estimate.
108 automobiles &times; 102 days &times; 101
gallons
= 1011 gallons
automobile ⋅ day
76. (a) Strategy There are 10,000 (104 ) half dollars in $5000. The mass of a half-dollar coin is about 10 grams, or
10−2 kilograms.
Solution Estimate the mass of the coins.
104 coins &times; 10−2 kg coin = 102 kg, or 100 kg .
(b) Strategy There are $1, 000, 000 $20 = 50, 000 twenty-dollar bills in $1,000,000. The mass of a twenty-dollar
bill is about 1 gram, or 10−3 kilograms.
Solution Estimate the mass of the bills.
50, 000 bills &times; 10−3 kg bill = 50 kg .
77. Strategy The SI base unit for mass is kg. Replace each quantity in W = mg with its SI base units.
Solution Find the SI unit for weight.
m
kg ⋅ m
kg ⋅ =
2
s
s2
78. Strategy It is given that T 2 ∝ r 3 . Divide the period of Mars by that of Venus.
Solution Compare the period of Mars to that of Venus.
2
TMars
3
⎛ r
⎞ 2
⎛ 2r
⎞
r3
2
, or TMars = ⎜⎜ Venus ⎟⎟
= Mars , so TMars
= ⎜⎜ Mars ⎟⎟ TVenus
2
3
TVenus
rVenus
⎝ rVenus ⎠
⎝ rVenus ⎠
32
TVenus = 23 2 TVenus ≈ 2.8TVenus .
79. Strategy $59,000,000,000 has a precision of 1 billion dollars; $100 has a precision of 100 dollars, so the net
worth is the same to one significant figure.
Solution Find the net worth.
$59, 000, 000, 000 − $100 = $59, 000, 000, 000
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80. Strategy There are about 103 hairs in a one-square-inch area of the average human head. An order-of-magnitude
estimate of the area of the average human head is 102 square inches.
Solution Calculate the estimate.
103 hairs in 2 &times; 102 in 2 = 105 hairs
81. (a) Strategy There are 7.0 leagues in one pace and 4.8 kilometers in one league.
Solution Find your speed in kilometers per hour.
120 paces 7.0 leagues 4.8 km 60 min
&times;
&times;
&times;
= 2.4 &times; 105 km h
1 min
1 pace
1 league
1h
(b) Strategy The circumference of the earth is approximately 40,000 km. The time it takes to march around the
Earth is found by dividing the distance by the speed.
Solution Find the time of travel.
1h
60 min
40,000 km &times;
&times;
= 10 min
1h
2.4 &times; 105 km
82. Strategy Use the fact that RB = 1.42 RA .
Solution Calculate the ratio of PB to PA .
V2
PB RB RA
RA
1
= 2 =
=
=
= 0.704
PA V
RB 1.42 RA 1.42
R
A
83. (a) Strategy Inspect the units of G, c, and h and use trial-and-error to find the correct combination of these
constants.
Solution Through a process of trial and error, we find that the only combination of G, c, and h that has the
hG
c5
dimensions of time is
.
(b) Strategy Substitute the values of the constants into the formula found in part (a).
Solution Find the time in seconds.
2
⎛
−34 kg⋅m ⎞ ⎛ 6.7 &times; 10−11 m3 ⎞
⎜ 6.6 &times; 10
⎟⎜
⎟
s
kg⋅s 2 ⎠
⎠⎝
== ⎝
= 1.3 &times; 10−43 s
5
8
c5
m
3.0 &times; 10 s
hG
(
)
84. Strategy The dimensions of L, g, and m are length, length per time squared, and mass, respectively. The period
has units of time, so T cannot depend upon m. (There are no other quantities with units of mass with which to
cancel the units of m.) Use a combination of L and g.
Solution The square root of L g has dimensions of time, so
T = C Lg , where C is a constant of proportionality .
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Chapter 1: Introduction
85. Strategy The dimensions of k and m are mass per time squared and mass, respectively. Dividing either quantity
by the other will eliminate the mass dimension.
Solution The square root of k m has dimensions of inverse time, which is correct for frequency.
So, f = k m . Find k.
f1 =
k
k
, so f12 =
, or k = m1 f12 .
m1
m1
Find the frequency of the chair with the 75-kg astronaut.
f2 =
k
=
m2
m1 f12
m2
= f1
m1
62 kg + 10.0 kg
= (0.50 s −1 )
= 0.46 s −1
m2
75 kg + 10.0 kg
86. Strategy Solutions will vary. One example follows:
The radius of the Earth is about 106 m. The area of a sphere is 4π r 2 , or about 101 ⋅ r 2 . The average depth of the
oceans is about 4 &times; 103 m. The oceans cover more than two-thirds of the Earth’s surface, but in this rough
estimation, we assume that oceans cover the entire Earth.
Solution Calculate an order-of-magnitude estimate of the volume of water contained in Earth’s oceans.
The surface area of the Earth is about 101 ⋅ (106 m)2 = 1013 m 2 ; therefore, the volume of water in the oceans is
about area &times; depth = (1013 m 2 )(4 &times; 103 m) = 4 &times; 1016 m3 ∼ 1016 m3 .
Total Mass of
Yeast Cells (g)
87. (a) Strategy Plot the data on a graph with
mass on the vertical axis and time on the
horizontal axis. Then, draw a best-fit
smooth curve.
Solution See the graph.
100.0
90.0
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
0.0
0.0
5.0 10.0 15.0 20.0 25.0
Time (h)
(b) Strategy Answers will vary. Estimate the value of the total mass that the graph appears to be approaching
asymptotically.
Solution The graph appears to be approaching asymptotically a maximum value of 100 g, so the carrying
capacity is about 100 g .
(c) Strategy Plot the data on a graph with the natural
logarithm of m m0 on the vertical axis and time on
the horizontal axis. Draw a line through the points and
find its slope to estimate the intrinsic growth rate.
m
ln m
0
2.0
1.0
Solution See the graph. From the plot of ln
m
m0
vs. t,
the slope r appears to be
1.8 − 0.0
1.8
r=
=
= 0.30 s −1 .
6.0 s − 0.0 s 6.0 s
0.0
0.0
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2.0
4.0
6.0 t (h)